Step: 1

Outlier of a data set is the data value that is far apart from the remaining values.

Step: 2

The height of each bar in the graph indicate the percentage loss of the company during the particular year.

Step: 3

From the graph, the loss percentages in the last 5 years are 5%, 4%, 6%, 1% and 5%.

Step: 4

Among the values 1% is far apart from the remaining ones.

Step: 5

So, the outlier loss percentage is 1%, which occured in the year 1998.

Correct Answer is : 1%, 1998

Step: 1

Outlier is the data value which is far apart from the remaining values.

Step: 2

In the data, there is no value which is far apart from the rest of the values.

[All values are in the same range.]

Step: 3

So, there is no outlier for the data.

Correct Answer is : No outlier

Step: 1

An outlier is the data item which is far apart from the remaining values.

Step: 2

No data value among the values is far apart from the others.

Step: 3

So, there is no outlier for the data set.

Correct Answer is : No outlier

Step: 1

Mean is a better measure of central tendency if there is no outlier for the data.

Step: 2

36 is the outlier of the data as it is far apart from other data values.

Step: 3

So, it may skew the central tendency.

Step: 4

As the outlier influences the mean, the median is the better measure of the central tendency.

Correct Answer is : Median

Step: 1

The outlier in a box-and-whisker plot is the data item, which is much higher or much lower than the other set of data items.

Step: 2

In the plot, the number 10 is much farther from the remaining set of data items.

Step: 3

So, the outlier in the plot is 10.

Correct Answer is : 10

Step: 1

Range is the difference between the maximum and minimum values of the data set.

Step: 2

Mode is used to find the occurrence of maximum times of a data value.

Step: 3

Mean is a better measure of central tendency if there is no outlier in the data.

Step: 4

27 is the outlier in the data as it is far apart from other values. So, it may skew the central tendency.

Step: 5

As the outlier influences the mean, the median is the better measure of the central tendency.

Correct Answer is : Median

Step: 1

Mean is a better measure of central tendency if there is no outlier for the data.

Step: 2

45 is the outlier of the data as it is far apart from other data values.

Step: 3

So, it may skew the central tendency.

Step: 4

As the outlier influences the mean, the median is the better measure of the central tendency.

Correct Answer is : Median

Step: 1

Arrange the weights of students in increasing order.

29, 36, 40, 45, 76

29, 36, 40, 45, 76

Step: 2

[Q 2 is the median, which is the middle value of the data set in the order.]

Step: 3

[Q 1 is the median of the data values less than Q 2.]

Step: 4

[Q 3 is the median of the data values greater than Q 2.]

Step: 5

Interquartile Range (IQR ) = Q 3 - Q 1

[Formula.]

Step: 6

= 60.5 - 32.5 = 28

Step: 7

To find the outlier, compute the cut-off points for outliers.

Step: 8

Lower fence = Q 1 - 1.5(IQR )

= 32.5 - 1.5(28) = - 9.5

= 32.5 - 1.5(28) = - 9.5

[Formula.]

Step: 9

Upper fence = Q 3 +1.5(IQR )

= 60.5 + 1.5(28) = 102.5

= 60.5 + 1.5(28) = 102.5

[Formula]

Step: 10

If a data value is less than the lower fence or greater than the upper fence, then it is considered as an outlier.

Step: 11

There are no values that are less than - 9.5 or greater than 102.5.

Step: 12

So, there is no outlier.

Correct Answer is : no outlier

3.5, 4.7, 5.3, 6.8, 4.2, 5.8, 12.5

Step: 1

Arrange the data values in increasing order.

3.5, 4.2, 4.7, 5.3, 5.8, 6.8, 12.5

3.5, 4.2, 4.7, 5.3, 5.8, 6.8, 12.5

Step: 2

[Q 2 is the median, which is the middle value of the data set in the order.]

Step: 3

[Q 1 is the median of the data values less than Q 2.]

Step: 4

[Q 3 is the median of the data values greater than Q 2.]

Step: 5

Interquartile Range (IQR ) = Q 3 - Q 1

= 6.8 - 4.2 = 2.6

= 6.8 - 4.2 = 2.6

[Formula.]

Step: 6

To find the outlier, compute the cut-off points for outliers.

Step: 7

Lower fence = Q 1 - 1.5(IQR )

= 4.2 - 1.5(2.6) = 0.3

= 4.2 - 1.5(2.6) = 0.3

[Formula.]

Step: 8

Upper fence = Q 3 + 1.5(IQR )

= 6.8 + 1.5(2.6) = 10.7

= 6.8 + 1.5(2.6) = 10.7

Step: 9

If a data value is less than the lower fence or greater than the upper fence, then it is considered as an outlier.

Step: 10

The value 12.5 is greater than 10.7.

Step: 11

So, 12.5 is the outlier.

Correct Answer is : 12.5

Step: 1

Arrange the number of days taken in increasing order.

7, 8, 12, 15, 17, 19, 23, 45

7, 8, 12, 15, 17, 19, 23, 45

Step: 2

[Q 2 is the median, which is the mean of the middle values of the data set in the order.]

Step: 3

[Q 1 is the median of the data values less than Q 2.]

Step: 4

[Q 3 is the median of the data values greater than Q 2.]

Step: 5

Interquartile Range(IQR ) = Q 3 - Q 1

= 21 - 10 = 11

= 21 - 10 = 11

[Formula.]

Step: 6

To find the outlier, compute the cut-off points for outliers.

Step: 7

Lower fence = Q 1 - 1.5(IQR)

= 10 - 1.5(11) = - 6.5

= 10 - 1.5(11) = - 6.5

[Formula]

Step: 8

Upper fence = Q 3 + 1.5(IQR )

= 21 + 1.5(11) = 37.5

= 21 + 1.5(11) = 37.5

[Formula]

Step: 9

If a data value is less than the lower fence or greater than the upper fence, then it is considered as an outlier.

Step: 10

The value 45 is greater than 37.5.

Step: 11

So, 45 is the outlier.

Correct Answer is : 45

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